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    CHAPTER OUTLINE

    1. Introduction

    2. Equilibrium of a deformable body

    3. Stress

    4. Average normal stress in an axially loaded bar

    5. Average shear stress

    6. Allowable stress

    7. Design of simple connections

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    Historical development

    Beginning of 17th century (Galileo) Early 18th century (Saint-Venant, Poisson, Lam

    and Navier)

    In recent times, with advanced mathematical and

    computer techniques, more complex problemscan be solved

    1.1 INTRODUCTION

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    1.2 EQUILIBRIUM OF A DEFORMABLE BODY

    External loads

    Surface forces

    Area of contact

    Concentrated force

    Linear distributed force Centroid C(or

    geometric center)

    Body force (e.g., weight)

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    Support reactions

    for 2D problems

    1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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    Equations of equilibrium

    For equilibrium balance of forces

    balance of moments

    Draw a free-body diagram to account for allforces acting on the body

    Apply the two equations to achieve equilibriumstate

    F= 0

    MO= 0

    1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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    Internal resultant loadings

    Define resultant force (FR) and moment (MRo) in 3D:

    Normal force, N

    Shear force, V

    Torsional moment or torque, T

    Bending moment, M

    1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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    Internal resultant loadings

    For coplanar loadings:

    Normal force, N

    Shear force, V

    Bending moment, M

    1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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    Internal resultant loadings

    For coplanar loadings:

    Apply Fx = 0 to solve for N

    Apply Fy = 0 to solve for V

    Apply MO = 0 to solve for M

    1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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    Procedure for Analysis

    Method of sections

    1. Choose segment to analyze

    2. Determine Support Reactions

    3. Draw free-body diagram for whole body

    4. Apply equations of equilibrium

    1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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    Procedure for analysis

    Free-body diagram

    1. Keep all external loadings in exact locations

    before sectioning

    2. Indicate unknown resultants, N, V, M, and Tat the section, normally at centroid Cof

    sectioned area

    3. Coplanar system of forces only include N, V,and M

    4. Establishx,y,zcoordinate axes with origin at

    centroid

    1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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    Procedure for analysis

    Equations of equilibrium

    1. Sum moments at section, about each

    coordinate axes where resultants act

    2. This will eliminate unknown forces Nand V,with direct solution for M(and T)

    3. Resultant force with negative value implies

    that assumed direction is opposite to thatshown on free-body diagram

    1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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    EXAMPLE 1.1

    Determine resultant loadings acting on cross

    section at Cof beam.

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    EXAMPLE 1.1 (SOLN)

    Support Reactions

    Consider segment CB

    Free-Body Diagram:

    Keep distributed loading exactly where it is on

    segment CBaftercutting the section.

    Replace it with a single resultant force,F.

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    EXAMPLE 1.1 (SOLN)

    Intensity (w) of loading at C(by proportion)

    w/6 m = (270 N/m)/9 m

    w= 180 N/m

    F= (180 N/m)(6 m) = 540 N

    Facts 1/3(6 m) = 2 m from C.

    Free-Body Diagram:

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    1 St

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    EXAMPLE 1.1 (SOLN)

    Equilibrium equations:

    Negative sign of Mcmeans it acts in the opposite

    direction to that shown below

    1 St

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    EXAMPLE 1.5

    Mass of pipe = 2 kg/m,

    subjected to vertical

    force of 50 N and couplemoment of 70 Nm at

    endA. It is fixed to the

    wall at C.

    Determine resultant internal loadings acting on cross

    section at Bof pipe.

    1 St

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    EXAMPLE 1.5 (SOLN)

    Support Reactions:

    Consider segmentAB,

    which does not involve

    support reactions at C.

    Free-Body Diagram:

    Need to find weight of

    each segment.

    1 St

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    EXAMPLE 1.5 (SOLN)

    WBD = (2 kg/m)(0.5 m)(9.81 N/kg)= 9.81 N

    WAD = (2 kg/m)(1.25 m)(9.81 N/kg)

    = 24.525 N

    1 St

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    EXAMPLE 1.5 (SOLN)

    Equilibrium equations:

    Fx= 0;

    Fy= 0;

    (FB)x= 0

    (FB

    )y= 0

    Fz= 0; (FB)z 9.81 N 24.525 N 50 N = 0(FB)z= 84.3 N

    1 St

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    EXAMPLE 1.5 (SOLN)

    Equilibrium Equations:

    (MB)x= 0;

    (Mc)x+ 70 Nm 50 N (0.5 m) 24.525 N (0.5 m)

    9.81 N (0.25m) = 0(MB)x= 30.3 Nm

    (MB)y= 0;

    (Mc)y+ 24.525 N (0.625m) + 50 N (1.25 m) = 0(MB)y= 77.8 Nm

    (MB)z= 0; (Mc)z= 0

    1 Stress

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    EXAMPLE 1.5 (SOLN)

    NB= (FB)y= 0

    VB= (0)2+ (84.3)2= 84.3 N

    TB= (MB)y= 77.8 Nm

    MB= (30.3)2+ (0)2= 30.3 Nm

    The direction of each moment is determinedusing the right-hand rule: positive moments

    (thumb) directed along positive coordinate axis

    Equilibrium Equations:

    1 Stress

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    1.3 STRESS

    Concept of stress

    To obtain distribution of force acting over a

    sectioned area

    Assumptions of material:

    1. It is continuous (uniform distribution of matter)2. It is cohesive (all portions are connected

    together)

    1 Stress

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    1.3 STRESS

    Concept of stress

    Consider Ain figure below

    Small finite force, Facts on A

    As A 0, F 0

    But stress (F /A) finite limit ()

    1 Stress

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    Normal stress

    Intensityof force, or force per unit area, acting

    normalto A

    Symbol used for normal stress,is (sigma)

    Tensile stress: normal force pulls or stretchesthe area element A

    Compressive stress: normal force pushes or

    compresses area element A

    1.3 STRESS

    z=lim

    A 0

    FzA

    1 Stress

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    Shear stress

    Intensityof force, or force per unit area, acting

    tangenttoA

    Symbol used for normal stress is (tau)

    1.3 STRESS

    zx=lim

    A 0

    FxA

    zy=lim

    A 0

    FyA

    1 Stress

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    General state of stress

    Figure shows the state of stressacting around a chosen point in abody

    Units (SI system)

    Newtons per square meter (N/m2)or apascal(1 Pa = 1 N/m2)

    kPa = 103

    N/m2

    (kilo-pascal) MPa = 106N/m2(mega-pascal)

    GPa = 109N/m2(giga-pascal)

    1.3 STRESS

    1 Stress

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    1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

    Examples of axially loaded bar

    Usually long and slender structural members

    Truss members, hangers, bolts

    Prismatic means all the cross sections are the same

    1 Stress

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    Assumptions

    1. Uniform deformation: Bar remains straight before

    and after load is applied, and cross section

    remains flat or plane during deformation

    2. In order for uniform deformation, force Pbeapplied along centroidal axis of cross section

    1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

    1 Stress

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    Average normal stress distribution

    = average normal stress at anypoint on cross sectional area

    P= internal resultant normal force

    A= x-sectional area of the bar

    1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

    FRz= Fxz dF = AdA

    P= A

    +

    P

    A=

    1 Stress

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    Equilibrium

    Consider vertical equilibrium of the element

    Fz= 0 (A) (A) = 0=

    Above analysis

    applies to members

    subjected to tensionor compression.

    1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

    1 Stress

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    Maximum average normal stress

    For problems where internal force Pand x-sectional A were constantalong the longitudinalaxis of the bar, normal stress =P/Ais alsoconstant

    If the bar is subjected to several external loadsalong its axis, change in x-sectional area mayoccur

    Thus, it is important to find the maximum

    average normal stress To determine that, we need to find the locationwhere ratioP/Ais a maximum

    1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

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    1 Stress

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    Procedure for Analysis

    Average normal stress Use equation of = P/Afor x-sectional area of a

    member when section subjected to internal

    resultant force P

    1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

    1 Stress

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    Procedure for Analysis

    Axially loaded members Internal Loading:

    Section memberperpendicularto its longitudinal

    axis at pt where normal stress is to be

    determined

    Draw free-body diagram

    Use equation of force equilibrium to obtain

    internal axial force Pat the section

    1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

    1. Stress

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    Procedure for Analysis

    Axially loaded members Average Normal Stress:

    Determine members x-sectional area at the

    section

    Compute average normal stress =P/A

    1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

    1. Stress

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    EXAMPLE 1.6

    Bar width = 35 mm, thickness = 10 mm

    Determine max. average normal stress in bar when

    subjected to loading shown.

    1. Stress

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    EXAMPLE 1.6 (SOLN)

    Internal loading

    Normal force diagram

    By inspection, largest

    loading area isBC,wherePBC= 30 kN

    1. Stress

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    EXAMPLE 1.6 (SOLN)

    Max. Average normal stress,

    BC=PBC

    A

    30(103) N

    (0.035 m)(0.010 m)= = 85.7 MPa

    1. Stress

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    EXAMPLE 1.8

    Specific weight st= 80 kN/m3

    Determine average compressive stress acting at

    pointsAandB.

    1. Stress

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    EXAMPLE 1.8 (SOLN)

    Internal loading

    Based on free-body diagram,

    weight of segmentABdetermined from

    Wst= stVst

    1. Stress

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    EXAMPLE 1.8 (SOLN)

    Average normal stress

    + Fz= 0; P Wst= 0

    P (80 kN/m3)(0.8 m)(0.2 m)2= 0

    P= 8.042 kN

    1. Stress

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    EXAMPLE 1.8 (SOLN)

    Average compressive stress

    Cross-sectional area at section is:

    A= (0.2)m2

    8.042 kN

    (0.2 m)2

    P

    A==

    = 64.0 kN/m2

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    1.5 AVERAGE SHEAR STRESS

    Shear stress is the stress component that act

    in the plane of the sectioned area.

    Consider a force Facting to the bar

    For rigid supports, and Fis large enough, bar

    will deform and fail along the planes identifiedbyABand CD

    Free-body diagram indicates that shear force,

    V=F/2be applied at both sections to ensure

    equilibrium

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    1.5 AVERAGE SHEAR STRESS

    Average shear stress over each

    section is:

    avg= average shear stress atsection, assumed to be sameat each pt on the section

    V= internal resultant shear force at

    section determined from

    equations of equilibrium

    A= area of section

    V

    Aavg=

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    1.5 AVERAGE SHEAR STRESS

    Case discussed above is example of simpleor

    direct shear Caused by the direct action of applied load F

    Occurs in various types of simple connections,e.g., bolts, pins, welded material

    1. Stress

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    Single shear

    Steel and wood joints shown below are

    examples of single-shear connections, also

    known as lap joints.

    Since we assume members are thin, there areno moments caused byF

    1.5 AVERAGE SHEAR STRESS

    1. Stress

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    Single shear

    For equilibrium, x-sectional area of bolt and

    bonding surface between the two members are

    subjected to single shear force, V=F

    The average shear stress equation can beapplied to determine average shear stress

    acting on colored section in (d).

    1.5 AVERAGE SHEAR STRESS

    1. Stress

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    1.5 AVERAGE SHEAR STRESS

    Double shear

    The joints shown below are examples of double-shear connections, often called double lap joints.

    For equilibrium, x-sectional area of bolt andbonding surface between two members

    subjected to double shear force, V=F/2 Apply average shear stress equation todetermine average shear stress acting oncolored section in (d).

    1. Stress

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    1.5 AVERAGE SHEAR STRESS

    Procedure for analysis

    Internal shear

    1. Section member at the pt where the avgis to be

    determined

    2. Draw free-body diagram3. Calculate the internal shear force V

    Average shear stress

    1. Determine sectioned areaA2. Compute average shear stress avg= V/A

    1. Stress

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    EXAMPLE 1.10

    Depth and thickness = 40 mm

    Determine average normal stress and average

    shear stress acting along (a) section planes a-a,

    and (b) section plane b-b.

    1. Stress

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    EXAMPLE 1.10 (SOLN)

    Part (a)

    Internal loadingBased on free-body diagram, Resultant loadingof axial force,P= 800 N

    1. Stress

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    EXAMPLE 1.10 (SOLN)

    Part (a)

    Average stressAverage normal stress,

    =P

    A

    800 N

    (0.04 m)(0.04 m) = 500 kPa=

    1. Stress

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    EXAMPLE 1.10 (SOLN)

    Part (a)

    Internal loadingNo shear stress on section, since shear force atsection is zero.

    avg= 0

    1. Stress

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    EXAMPLE 1.10 (SOLN)

    Part (b)

    Internal loading

    +

    Fx= 0; 800 N +Nsin 60+ Vcos 60 = 0+

    Fy= 0; Vsin 60 Ncos 60= 0

    1. Stress

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    EXAMPLE 1.10 (SOLN)

    Part (b)

    Internal loading

    Or directly usingx,yaxes,

    Fx= 0;

    Fy= 0;

    +

    +

    N 800 N cos 30= 0

    V 800 N sin 30= 0

    1. Stress

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    EXAMPLE 1.10 (SOLN)

    Part (b)

    Average normal stress

    =N

    A

    692.8 N

    (0.04 m)(0.04 m/sin 60)= 375 kPa=

    1. Stress

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    EXAMPLE 1.10 (SOLN)

    Part (b)

    Average shear stress

    avg =V

    A

    400 N

    (0.04 m)(0.04 m/sin 60)= 217 kPa=

    Stress distribution shown below

    1. Stress

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    1.6 ALLOWABLE STRESS

    When designing a structural member or

    mechanical element, the stress in it must berestricted to safe level

    Choose an allowable load that is less than theload the member can fully support

    One method used is the factor of safety (F.S.)

    F.S.=Ffail

    Fallow

    1. Stress

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    1.6 ALLOWABLE STRESS

    If load applied is linearly related to stress

    developed within member, then F.S. can alsobe expressed as:

    F.S.=fail

    allowF.S.=

    fail

    allow

    In all the equations, F.S. is chosen to be greater than 1,to avoid potential for failure

    Specific values will depend on types of material usedand its intended purpose

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    1.7 DESIGN OF SIMPLE CONNECTIONS

    To determine area of section subjected to a

    normal force, use

    A=P

    allow

    A=

    V

    allow

    To determine area of section subjected to a shearforce, use

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    1.7 DESIGN OF SIMPLE CONNECTIONS

    Cross-sectional area of a tension member

    Condition:

    The force has a line of action that passes

    through the centroid of the cross section.

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    1.7 DESIGN OF SIMPLE CONNECTIONS

    Cross-sectional area of a connecter subjected to

    shear

    Assumption:

    If bolt is loose or clamping force of bolt is unknown,assume frictional force between plates to be

    negligible.

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    Assumptions:

    1. (b)allowof concrete